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Question

If 2x3+ax2+bx6 has (x1) as a factor and leaves a remainder 2 when divided by (x2), find the value of 'a' and 'b' respectively.

A
8,12
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B
8,12
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C
4,10
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D
4,10
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Solution

The correct option is A 8,12
Let f(x)=2x3+ax2+bx6

Since (x1) is the factor of f(x), therefore, by factor theorem, f(1)=0 that is:

f(1)=2(1)3+a(1)2+(b×1)60=(2×1)+(a×1)+b60=2+a+b6a+b=62a+b=4.......(1)

Also, it is given that when f(x) is divided by (x2), it leaves remainder 2, therefore, by remainder theorem, f(2)=2 that is:

f(2)=2(2)3+a(2)2+(b×2)62=(2×8)+(a×4)+2b62=16+4a+2b64a+2b=2+6164a+2b=8.......(2)

Multiply equation 1 by 2 as follows:

2(a+b)=2×42a+2b=8.....(3)

Now, subtract equation 3 from equation 2 as shown below:

(4a2a)+(2b2b)=882a=16a=162a=8

Substituting the value of a in equation 1, we have:

a+b=48+b=4b=4+8b=12

Hence, a=8 and b=12.

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